Let $\epsilon^{A}_B$ be the map $\eta_{\frac{B}{A},A,B}^{-1}(1_{\frac{B}{A}}) $, or in other words, let $\epsilon^{A}$ be the counit of the adjunction $? \otimes_A\dashv \frac{??}{A}$, and similarly for $B,C$. Then the inverse of $\eta_{A,B,C}$ is the map $g\mapsto \epsilon^{B}_C \circ (g\otimes B)$. Moreover, the inner composition
$$\frac{C}{B}\otimes\frac{B}{A}\to\frac{C}{A}$$
is the image of the composite
$$\frac{C}{B}\otimes\frac{B}{A}\otimes A\stackrel{\frac{C}{B}\otimes \epsilon_{B}^{A}}{\longrightarrow} \frac{C}{B}\otimes B \stackrel{ \epsilon_{C}^B}{\longrightarrow} C$$
under the bijection $\eta_{\frac{C}{B}\otimes\frac{B}{A},A,C}$.
As a consequence, and because of the naturality of $\eta$, we find that
\begin{align}\bullet_{A,B,C}\circ (\widehat{\psi}\otimes\widehat{\varphi}) & = \eta_{\frac{C}{B}\otimes\frac{B}{A},A,C}\left(\epsilon^B_C\circ \left(\frac{C}{B}\otimes\epsilon^{A}_B\right) \right) \circ (\widehat{\psi}\otimes \widehat{\varphi})\\
& = \eta_{I\otimes I,A,C}\left(\epsilon^B_C\circ \left(\frac{C}{B}\otimes\epsilon^{A}_B\right) \circ (\widehat{\psi}\otimes \widehat{\varphi}\otimes A)\right) \\
& = \eta_{I\otimes I,A,C}\left(\epsilon^B_C\circ \left(\widehat{\psi}\otimes (\epsilon^{A}_B \circ ( \widehat{\varphi}\otimes A))\right) \right) \\
& = \eta_{I\otimes I,A,C}\left(\epsilon^B_C \circ \left(\widehat{\psi}\otimes (\varphi\circ \lambda_A) \right) \right) \\ &
= \eta_{I,A,C}\left(\epsilon^B_C \circ (\widehat{\psi}\otimes B) \circ (I\otimes (\varphi\circ \lambda_A)) \right) \\
& = \eta_{I\otimes I,A,C}\left(\psi\circ \lambda_B \circ (I\otimes (\varphi\circ \lambda_A)) \right)\\
& = \eta_{I\otimes I,A,C}\left(\psi\circ \varphi\circ \lambda_A \circ \lambda_{I\otimes A} \right) \\
& = \eta_{I\otimes I,A,C}\left((\psi\circ \varphi)\circ \lambda_A \circ (\lambda_{I}\otimes A) \right)\\
& = \eta_{I,A,C}\left((\psi\circ \varphi)\circ \lambda_A \right)\circ \lambda_{I} \\ & = \widehat{\psi\circ \varphi} \circ \lambda_I\end{align}
(with some associators missing, but it should work).
Best Answer
I will refer to https://ncatlab.org/nlab/show/internal+category for the axioms of an internal category, as I don't want to format that many diagrams :)
As there is only one functor from any category into the terminal category by definition, the diagrams for the laws of specifying the source and target of the identities are trivially fulfilled, as are the laws specifying the source and target of a composition.
However, and this is a critical point, the associativity and the unit law are NOT fulfilled. The reason for that, to spell it out, is that $ X \times * \neq X \neq * \times X$ and $X \times (Y \times Z) \neq (X \times Y) \times Y$ in general, but they are only canonically isomorphic. Maybe there is some notion of internal category in higher category theory that allows for this additional freedom (i.e. that these diagrams need only commute up to coherent isomorphisms), but at least with the standard definition your statement seems to not be true.